Download Chapter 10 Solids & Liquids continued

Chapter 10
Solids & Liquids
continued
10.3 Pascal’s Principle
PASCAL’S PRINCIPLE
Any change in the pressure applied
to a completely enclosed fluid is transmitted
undiminished to all parts of the fluid and
enclosing walls.
F2
F1
P2 = ; P1 =
A2
A1
Assume weight of fluid
in the tube is negligible
P2 = P1 + ρ gh
P >> ρ gh
P2 = P1
Small ratio
F2 F1
= ⇒
A2 A1
⎛ A1 ⎞
F1 = F2 ⎜ ⎟
⎝ A2 ⎠
10.4 Archimedes’ Principle
Liquid density
m
ρ (rho) =
V
P2 = P1 + ρ gh
Pressure grows linearly
with depth.
F1 =
F2 =
Buoyant Force, FB
P1
FB = F2 + (− F1 )
P2
= ρ ghA
= P2 A − P1 A = ( P2 − P1 ) A
=
ρV

since P2 − P1 = ρ gh
g
and V = hA
mass of
displaced fluid
Buoyant force = Weight of displaced fluid
FB = ρVg
10.4 Archimedes’ Principle
ARCHIMEDES’ PRINCIPLE
Any fluid applies a buoyant force to an object that is partially
or completely immersed in it; the magnitude of the buoyant
force equals the weight of the fluid that the object displaces:
CORROLARY
If an object is floating then the
magnitude of the buoyant force
is equal to the magnitude of its
weight.
10.4 Archimedes’ Principle
Any fluid applies a buoyant force to an object that is partially
or completely immersed in it; the magnitude of the buoyant
force equals the weight of the fluid that the object displaces:
FB = ρ WaterVg
W = ρObjectVg
FB = W − T
(Net force = 0)
FB
T
= 1−
W
W
ρ WaterVg
T
= 1−
ρObjectVg
W
ρObject
ρ Water
=
(1− T / W )
10.4 Archimedes’ Principle
Use Archimedes Principle to measure the density of Aluminum
ρObject
ρ Water
=
(1− T / W )
T 4.5N
~
= 0.53
W 8.5N
1000 kg/m 3
=
= 2100 kg/m 3
1− 0.53
10.4 Archimedes’ Principle
Example: A Swimming Raft
The raft is made of solid square
pinewood.
a) Determine whether the raft floats
in water and if so,
b) how much of the raft is beneath
the surface.
10.4 Archimedes’ Principle
Raft has these properties
If Wraft < FB submerged, raft floats
Wraft
Weight of the Raft
= mraft g = ρraftVraft g
(
)(
Vraft = ( 4.0 ) ( 4.0 ) ( 0.30 ) m 3 = 4.8 m 3
ρraft = 550 kg/m 3
)(
= 550kg m 3 4.8m 3 9.80m s 2
)
= 26000 N
Buoyant force totally submerged
FB = ρ Vg = ρWaterVRaft g
(
)(
)(
= 1000kg m 3 4.8m 3 9.80m s 2
)
= 47000 N
Wraft < FB submerged, so Raft floats
Only part of the raft is above water
10.4 Archimedes’ Principle
Raft has these properties
Vraft = ( 4.0 ) ( 4.0 ) ( 0.30 ) m 3 = 4.8 m 3
ρraft = 550 kg/m 3
An object will float in a fluid if its
density is less than that of the fluid !
ρ Raft = 550kg/m 3 < 1000kg/m 3 = ρWater
So, how much of the raft is beneath
the surface?
Only part of the raft is above water
10.4 Archimedes’ Principle
How much of raft below water?
For a floating object
FB = WRaft = 26000 N
FB is weight of displaced water
FB = ρ Water gVWater (displaced)
= ρ Water g( ARaft h)
WRaft
h=
(height below surface)
ρ Water gARaft
=
26000N
(1000kg m )(9.80m s )(16.0 m )
= 0.17 m
3
2
2
10.4 Archimedes’ Principle
In General
What fraction of full height H
is below the surface ?
WRaft
h=
ρ Water gARaft
ρ Raft gARaft H
=
ρ Water gARaft
ρ Raft
h
=
fraction below surface
H ρ Water
Fraction below surface is just the ratio of the density
of floating object to the density of the fluid.
H = 0.30 m
10.5 Fluids in Motion
In steady flow the velocity of the fluid particles at any point is constant
as time passes.
Unsteady flow exists whenever the velocity of the fluid particles at a
point changes as time passes.
Turbulent flow is an extreme kind of unsteady flow in which the velocity
of the fluid particles at a point change erratically in both magnitude and
direction.
Fluid flow can be compressible or incompressible. Most liquids are
nearly incompressible.
Fluid flow can be viscous or nonviscous.
An incompressible, nonviscous fluid is called an ideal fluid.
10.5 Fluids in Motion
When the flow is steady, streamlines
are often used to represent
the trajectories of the fluid particles.
The mass of fluid per second that flows
through a tube is called
the mass flow rate.
10.5 The Equation of Continuity
EQUATION OF CONTINUITY
The mass flow rate has the same value at every position along a
tube that has a single entry and a single exit for fluid flow.
SI Unit of Mass Flow Rate: kg/s
Δm2
= ρ2 A2 v2
Δt
Incompressible fluid:
Volume flow rate Q:
ρ1 = ρ2
10.5 The Equation of Continuity
Example: A Garden Hose
A garden hose has an unobstructed opening
with a cross sectional area of 2.85x10-4m2.
It fills a bucket with a volume of 8.00x10-3m3
in 30 seconds.
Find the speed of the water that leaves the hose
through (a) the unobstructed opening and (b) an obstructed
opening with half as much area.
a) Q = Av
Q
v= =
A
b)
(
)
8.00 × 10−3 m 3 / ( 30.0 s )
2.85 × 10 m
-4
2
= 0.936m s
A1v1 = A2 v2
A1
v2 =
v1 = ( 2 ) ( 0.936m s ) = 1.87 m s
A2
10.5 Bernoulli’s Equation
The fluid accelerates toward the
lower pressure regions.
According to the pressure-depth
relationship, the pressure is lower
at higher levels, provided the area
of the pipe does not change.
Apply Work-Energy theorem
to determine relationship between
pressure, height, velocity, of the fluid.
10.5 Bernoulli’s Equation
Work done by tiny pressure “piston”
WΔP =
(∑ F ) s = ( ΔF ) s = ( ΔPA) s;
V = As
Work (NC) done by pressure difference from 2 to 1
WNC = ( P2 − P1 )V
E1 = 12 mv12 + mgy1
E2 = 12 mv22 + mgy2
WNC = E1 − E2 =
(
1
2
) (
mv12 + mgy1 −
1
2
mv22 + mgy2
)
10.5 Bernoulli’s Equation
WNC = ( P2 − P1 )V
WNC = E1 − E2 =
(
1
2
) (
mv + mgy1 −
2
1
1
2
mv + mgy2
2
2
)
NC Work yields a
total Energy change.
Equating the two expressions for the work done,
( P − P )V = (
2
1
(P − P) = (
2
1
1
2
1
2
) (
mv12 + mgy1 −
) (
ρ v12 + ρ gy1 −
1
2
1
2
mv22 + mgy2
ρ v22 + ρ gy2
)
m = ρV
)
Rearrange to obtain Bernoulli's Equation
BERNOULLI’S EQUATION
In steady flow of a nonviscous, incompressible fluid, the pressure, the
fluid speed, and the elevation at two points are related by:
P1 + 12 ρ v12 + ρ gy1 = P2 + 12 ρ v22 + ρ gy2
10.5 Applications of Bernoulli’s Equation
Conceptual Example: Tarpaulins and Bernoulli’s Equation
When the truck is stationary, the
tarpaulin lies flat, but it bulges outward
when the truck is speeding down
the highway.
Account for this behavior.
Relative to moving truck
v1 = 0 under the tarp
Bernoulli’s Equation
P1 + ρ v + ρ gy1 = P2 + ρ v + ρ gy2
1
2
2
1
1
2
P1 = P2 + 12 ρ v22
P1 > P2
2
2
v2 air flow over top
10.5 Applications of Bernoulli’s Equation
10.5 Applications of Bernoulli’s Equation
Example: Efflux Speed
The tank is open to the atmosphere at
the top. Find and expression for the speed
of the liquid leaving the pipe at
the bottom.
P1 = P2 = Patmosphere (1× 105 N/m 2 )
v2 = 0,
y2 = h,
y1 = 0
P1 + 12 ρ v12 + ρ gy1 = P2 + 12 ρ v22 + ρ gy2
1
2
ρ v12 = ρ gh
v1 = 2gh
10.6 Viscous Flow
Flow of an ideal fluid.
Flow of a viscous fluid.
FORCE NEEDED TO MOVE A LAYER OF VISCOUS FLUID WITH
CONSTANT VELOCITY
The magnitude of the tangential force required to move a fluid
layer at a constant speed is given by:
η Av
F=
y
η , is the coefficient of viscosity
SI Unit: Pa ⋅s; 1 poise (P) = 0.1 Pa ⋅s
POISEUILLE’S LAW (flow of viscous fluid)
The volume flow rate is given by:
Pressure drop in a
straight uniform
diamater pipe.
11.11 Viscous Flow
Example: Giving and Injection
A syringe is filled with a solution whose
viscosity is 1.5x10-3 Pa·s. The internal
radius of the needle is 4.0x10-4m.
The gauge pressure in the vein is 1900 Pa.
What force must be applied to the plunger,
so that 1.0x10-6m3 of fluid can be injected
in 3.0 s?
8η LQ
P2 − P1 =
π R4
8 1.5 × 10−3 Pa ⋅s ( 0.025 m ) 1.0 × 10−6 m 3 3.0 s
=
= 1200 Pa
4
π 4.0 × 10-4 m
(
)
(
(
)
P2 = (1200 + P1 ) Pa = (1200 + 1900 ) Pa = 3100 Pa
(
)
F = P2 A = ( 3100 Pa ) 8.0 × 10−5 m 2 = 0.25N
)